Em tham khảo nha :

\(\begin{array}{l}
a)\\
2KCl{O_3} \to 2KCl + 3{O_2}\\
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
{m_{KCl{O_3}}} = \dfrac{{48,2 \times 50,83}}{{100}} = 24,5g\\
{m_{KMn{O_4}}} = 48,2 - 24,5 = 23,7g\\
{n_{KCl{O_3}}} = \dfrac{{24,5}}{{122,5}} = 0,2mol\\
{n_{KMn{O_4}}} = \dfrac{{23,7}}{{158}} = 0,15mol\\
{n_{{O_2}}} = \dfrac{3}{2}{n_{KCl{O_3}}} + \dfrac{{{n_{KMn{O_4}}}}}{2} = 0,375mol\\
{V_{{O_2}}} = 0,375 \times 22,4 = 8,4l\\
b)\\
3Fe + 2{O_2} \to F{e_3}{O_4}\\
{n_{Fe}} = \dfrac{{25,2}}{{56}} = 0,45mol\\
\dfrac{{0,45}}{3} < \dfrac{{0,375}}{2} \Rightarrow {O_2}\text{ dư}\\
{n_{F{e_3}{O_4}}} = \dfrac{{{n_{Fe}}}}{3} = 0,15mol\\
{m_{F{e_3}{O_4}}} = 0,15 \times 232 = 34,8g
\end{array}\)